Inflection Points Of F(x) = X^2 * E^(16x) Finding C And D

Hey guys! Let's dive into a fascinating mathematical journey where we'll explore the inflection points of the function f(x) = x²e¹⁶ˣ. This is a classic calculus problem that involves finding where the concavity of the function changes. Think of it like figuring out where a curve transitions from being 'smiling' to 'frowning' or vice versa. It's super cool stuff, so let's get started!

Understanding Inflection Points

Before we jump into the calculations, let's quickly recap what inflection points are. Inflection points are those special spots on a curve where the concavity changes. Concavity, in simple terms, describes the direction in which a curve is bending. If a curve is concave up, it looks like a cup opening upwards (think of a smile 😃). If it's concave down, it looks like a cup opening downwards (think of a frown 🙁).

To find these inflection points, we need to use calculus! Specifically, we need to find the second derivative of our function, f(x). The second derivative tells us about the rate of change of the slope, which is directly related to concavity. We'll set the second derivative equal to zero and solve for x. These x values are our potential inflection points. However, we also need to check that the concavity actually changes at these points. We can do this by examining the sign of the second derivative on either side of these points.

So, in a nutshell, finding inflection points involves these key steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for x. These are our candidate inflection points.
3. Check the concavity (sign of the second derivative) on either side of each candidate point. If the concavity changes, then it's a true inflection point.

Step-by-Step Calculation of Inflection Points for f(x) = x²e¹⁶ˣ

Okay, enough talk! Let's get our hands dirty with some actual math. We're going to find the inflection points of f(x) = x²e¹⁶ˣ step by step. Trust me, it's not as scary as it looks!

1. Finding the First Derivative, f'(x)

The first thing we need to do is find the first derivative of f(x). Remember, the derivative tells us about the slope of the function at any given point. To find the derivative of f(x) = x²e¹⁶ˣ, we'll need to use the product rule. The product rule states that if we have a function that's the product of two other functions, say u(x) and v(x), then the derivative of the product is:

(uv)' = u'v + uv'

In our case, we can consider u(x) = x² and v(x) = e¹⁶ˣ. So, let's find their derivatives:

• u'(x) = 2x
• v'(x) = 16e¹⁶ˣ (using the chain rule)

Now, we can plug these into the product rule formula:

f'(x) = (2x)(e¹⁶ˣ) + (x²)(16e¹⁶ˣ)

Let's simplify this a bit by factoring out the common term 2xe¹⁶ˣ:

f'(x) = 2xe¹⁶ˣ(1 + 8x)

Great! We've found the first derivative. This will be useful later, but for now, we need to move on to the second derivative.

2. Finding the Second Derivative, f''(x)

Now comes the fun part – finding the second derivative, f''(x). This is the derivative of the first derivative, f'(x). Remember, the second derivative tells us about the concavity of the function. Since our first derivative, f'(x) = 2xe¹⁶ˣ(1 + 8x), is also a product of functions, we'll need to use the product rule again. But this time, we have three factors! Don't worry, the product rule can be extended to handle this. One way to approach this is to consider 2x as one function, and e¹⁶ˣ(1 + 8x) as another function, apply the product rule, and then apply the product rule again to the second part. Alternatively, we can distribute first to get two terms and then apply the product rule to each term individually.

Let's distribute first for simplicity. So, f'(x) = 2xe¹⁶ˣ + 16x²e¹⁶ˣ. Now, we apply the product rule to each term separately:

For the first term, 2xe¹⁶ˣ: u(x) = 2x, v(x) = e¹⁶ˣ, u'(x) = 2, v'(x) = 16e¹⁶ˣ. Applying the product rule: 2e¹⁶ˣ + 32xe¹⁶ˣ.

For the second term, 16x²e¹⁶ˣ: u(x) = 16x², v(x) = e¹⁶ˣ, u'(x) = 32x, v'(x) = 16e¹⁶ˣ. Applying the product rule: 32xe¹⁶ˣ + 256x²e¹⁶ˣ.

Now, combining both results:

f''(x) = (2e¹⁶ˣ + 32xe¹⁶ˣ) + (32xe¹⁶ˣ + 256x²e¹⁶ˣ)

Simplifying by combining like terms:

f''(x) = 2e¹⁶ˣ + 64xe¹⁶ˣ + 256x²e¹⁶ˣ

Let's factor out the common term 2e¹⁶ˣ to make it look even nicer:

f''(x) = 2e¹⁶ˣ(1 + 32x + 128x²)

Woohoo! We've found the second derivative. This is the key to finding our inflection points.

3. Finding Potential Inflection Points by Setting f''(x) = 0

Remember, inflection points occur where the concavity changes, which means the second derivative is either zero or undefined. In our case, f''(x) = 2e¹⁶ˣ(1 + 32x + 128x²). Since e¹⁶ˣ is never zero, we only need to worry about the quadratic part:

1 + 32x + 128x² = 0

This is a quadratic equation, and we can solve it using the quadratic formula. The quadratic formula states that for an equation of the form ax² + bx + c = 0, the solutions for x are:

x = (-b ± √(b² - 4ac)) / (2a)

In our case, a = 128, b = 32, and c = 1. Let's plug these values into the quadratic formula:

x = (-32 ± √(32² - 4 * 128 * 1)) / (2 * 128)

Let's simplify this:

x = (-32 ± √(1024 - 512)) / 256 x = (-32 ± √512) / 256 x = (-32 ± 16√2) / 256

We can simplify this further by dividing both the numerator and denominator by 32:

x = (-1 ± √2 / 2) / 8

So, we have two potential inflection points:

• x₁ = (-1 - √2 / 2) / 8 which is approximately -0.213
• x₂ = (-1 + √2 / 2) / 8 which is approximately -0.037

Let's call these C and D, where C ≤ D. So:

• C = (-1 - √2 / 2) / 8 ≈ -0.213
• D = (-1 + √2 / 2) / 8 ≈ -0.037

4. Verifying Inflection Points by Checking Concavity

We've found two potential inflection points, but we need to make sure that the concavity actually changes at these points. To do this, we'll pick test values in the intervals determined by C and D and plug them into the second derivative, f''(x). The sign of f''(x) will tell us about the concavity in that interval.

Our intervals are:

• x < C ≈ -0.213
• C < x < D ≈ -0.213 < x < -0.037
• x > D ≈ -0.037

Let's pick some test values:

• x = -1 (in the interval x < C)
• x = -0.1 (in the interval C < x < D)
• x = 0 (in the interval x > D)

Now, let's plug these values into f''(x) = 2e¹⁶ˣ(1 + 32x + 128x²):

• f''(-1) = 2e⁻¹⁶(1 - 32 + 128) = 2e⁻¹⁶(97) > 0 (Concave Up)
• f''(-0.1) = 2e⁻¹·⁶(1 - 3.2 + 1.28) = 2e⁻¹·⁶(-0.92) < 0 (Concave Down)
• f''(0) = 2e⁰(1 + 0 + 0) = 2 > 0 (Concave Up)

As we can see, the concavity changes at both x = C and x = D. The function is concave up for x < C, concave down for C < x < D, and concave up again for x > D. This confirms that both C and D are indeed inflection points.

The Grand Finale: Our Inflection Points

Alright guys, we made it! We've successfully navigated the world of derivatives and concavity to find the inflection points of f(x) = x²e¹⁶ˣ. We found that:

• C = (-1 - √2 / 2) / 8 ≈ -0.213
• D = (-1 + √2 / 2) / 8 ≈ -0.037

These are the x-coordinates where the concavity of the function changes. If you wanted to find the y-coordinates of these inflection points, you'd simply plug these x-values back into the original function, f(x) = x²e¹⁶ˣ.

Why Are Inflection Points Important?

You might be wondering,